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This was one of my exam questions and the answer is apparently no.

Can someone explain why because I don't understand.

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  • $\begingroup$ @Jut, please do not use the comments here to advertise one of your other posts. That is not an appropriate use of comments. $\endgroup$
    – D.W.
    Dec 20, 2022 at 6:14

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There exist languages which are Non-RE and their complement is also Non-RE, let $L$ be such a language we can say that

$L \cup L^{c} = \Sigma^{*}$

Here both languages are Non-RE but their union is $\Sigma^{*}$ which is regular as well as RE also.

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